This is a method I tried of working out the cartesian form of a hyperbola using its definition as the cross section of a vertical cone?
I started by noting that the width of the hyperbola is the sine of a circle at some point within the cone. The radius of the circle is proportional to the height of the cone, which is the length $y$ down the hyperbola plus some constant $a$. Therefore the radius of the circle is $k(y+a)$.
The centre of the hyperbola is always the same distance from the centre of the cone at a given height $y$, at $ka$. So cosine of the angle between the middle and edge of the hyperbola at some height $y$ is $\frac{ka}{k(y+a)} = \frac{1}{1+\frac{y}{a}}$. So the width of the hyperbola $x$ at height $y$ is $x =k(y+a)\sqrt{1-\frac{1}{(1+\frac{y}{a})^2}}$ by relating the sines to cosines. The resulting equation is $x^2 = k^2(y+a)^2(1-\frac{1}{(1+\frac{y}{a})^2})$,
This simplifies to $x^2 = k^2(y^2 + 2ya)$. Is this a hyperbolic curve. How do I rearrange it to the standard form?
I understand that I'm not being very clear but I can't find a software on which to draw the problem properly. Anyway, I hope that the line of reasoning has come across.
Best Answer
$$ \eqalign{ & x^{\,2} = k^{\,2} \left( {y + a} \right)^{\,2} \left( {1 - {1 \over {\left( {1 + y/a} \right)^{\,2} }}} \right) = k^{\,2} a^{\,2} \left( {1 + y/a} \right)^{\,2} \left( {{{\left( {1 + y/a} \right)^{\,2} - 1} \over {\left( {1 + y/a} \right)^{\,2} }}} \right) = \cr & = k^{\,2} a^{\,2} \left( {\left( {1 + y/a} \right)^{\,2} - 1} \right) = k^{\,2} \left( {\left( {y + a} \right)^{\,2} - a^{\,2} } \right) \cr} $$ that is $$ {{\left( {y + a} \right)^{\,2} } \over {a^{\,2} }} - {{x^{\,2} } \over {\left( {ka} \right)^{\,2} }} = 1 $$ which looks correct in the reference system you adopted: $$ \left\{ \matrix{ x = 0\quad \to \quad y = - a \pm a \hfill \cr x = \pm ka\quad \to \quad y = - a \pm \sqrt 2 a \hfill \cr} \right. $$